JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
Let Sₙ be the sum of the first n terms of an arithmetic progression. If S₃ₙ = 3S₂ₙ, then the value of S₄ₙ/S₂ₙ is :
- (a) 6
- (b) 4
- (c) 2
- (d) 8
Answer and solution
Answer (a) 6
Solution
Maths sir · 3 steps
Write the AP sum as Sₖ = Pk² + Qk; the condition ties Q to P and the rest cancels.
- Quadratic form. Sₖ = k/2(2a + (k-1)d) = d/2k² + (a - d/2)k. So Sₖ = Pk² + Qk with constants P = d/2 and Q = a - d/2. There is no constant term.
- Use the condition. S₃ₙ = 9Pn² + 3Qn and 3S₂ₙ = 12Pn² + 6Qn. Setting them equal gives 3Pn² + 3Qn = 0, so Q = -Pn.
S₄ₙ/S₂ₙ = (16Pn² + 4Qn)/(4Pn² + 2Qn) = (16Pn² - 4Pn²)/(4Pn² - 2Pn²) = 12Pn²/2Pn²
- So. Pn² cancels and the ratio is 6, option (a).
The idea to keep. For any AP, Sₖ = Pk² + Qk with no constant term. When a question relates sums like S₂ₙ, S₃ₙ, S₄ₙ, plug in this form, find Q in terms of P, and let P cancel.
Answer from the NTA final answer key. Solution by PrepWiser.